If z is a complex number,then the curves |z|= | vedclass

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(C) Given the equations of the curves:$|z|=1 \Rightarrow x^2+y^2=1$ $(i)$$|z-2|=1 \Rightarrow (x-2)^2+y^2=1$ $(ii)$$|z-1|=0 \Rightarrow (x-1)^2+y^2=0$

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$(1,0)$

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(C) Given the equations of the curves:$|z|=1 \Rightarrow x^2+y^2=1$ $(i)$$|z-2|=1 \Rightarrow (x-2)^2+y^2=1$ $(ii)$$|z-1|=0 \Rightarrow (x-1)^2+y^2=0$ $(iii)$Solving equations $(i)$ and $(ii)$:$(x-2)^2+y^2 = x^2+y^2$$(x-2)^2 = x^2$$x^2-4x+4 = x^2$$4x = 4 \Rightarrow x=1$Substituting $x=1$ into equation $(i)$:$1^2+y^2=1$ $\Rightarrow y^2=0$ $\Rightarrow y=0$So,the intersection point of $(i)$ and $(ii)$ is $(1,0)$.Now,check if $(1,0)$ satisfies equation $(iii)$:$(1-1)^2+0^2 = 0^2+0^2 = 0$Since it satisfies all three equations,the common point is $(1,0)$.

If $z$ is a complex number,then the curves $|z|=1$,$|z-2|=1$,and $|z-1|=0$ have a common point at

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